16-Civ-A2 Elementary Structural Design · May 2013

Question 3 of 7: A3 — Maximum Factored Bracket Load on a CHS Column

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2013 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, CLOSED BOOK (handbooks and textbooks permitted, no notes). Seven questions are printed: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete) and Part C (C1, timber). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here, because the set is a study resource. All loads shown on the figures are unfactored; NBCC load combination 1.25D + 1.5L is applied throughout. Solutions are to CSA-S16 (steel), CAN/CSA-A23.3 (concrete) and CSA-O86 (timber), as the paper directs.

Reference texts.

CSA S16, Design of Steel Structures — Clauses 11 (classification), 13.3 (compression), 13.5 (bending), 13.8 (beam-columns), 13.13 (welds).
CISC, Handbook of Steel Construction — section-property tables for C-shapes, W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial load) and 11 (shear, general and simplified methods).
Brzev & Pao, Reinforced Concrete Design: A Practical Approach — Chapters 3–6 and 9.
CSA O86, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B and 6.4.5.
Canadian Wood Council, Wood Design Manual — beam selection and modification factors.

Check — assumptions carried through this paper. (1) The 60 kN, 50 kN, 40 kN, 200 kN, 250 kN and 300 kN forces on Figures A1–B3 are specified live loads, so a load factor of 1.5 is used; concrete self-weight is factored at 1.25. (2) Steel members are assumed continuously laterally supported unless an unbraced length is printed — no unbraced lengths appear on this paper. (3) In Figure B2 the "65 typical" dimension is taken as the distance from the tension face to the centroid of the bottom bars, and the 15M stirrups are assumed at 300 mm on centre in each stem (the figure gives no spacing). (4) CSA O86 Table 6.3.1B Beam-and-Stringer specified strengths for D.Fir-L No.1 are used in C1.

Question 3: A3 — Maximum Factored Bracket Load on a CHS Column (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Circular hollow section, $D = 406.4$ mm, $t = 9.53$ mm, G40.21 350W Class H ($F_y = 350$ MPa, $E = 200\,000$ MPa). Height $L = 6.0$ m, hinged at the top and fixed at the base. Vertical bracket load $P_F$ at eccentricity $e = 0.8$ m, applied at the top of the column.

Find. The largest factored bracket load $P_F$ the column can carry as a beam-column.

Figure A3 (from the question text) — 6 m circular hollow column, laterally supported and free to rotate at the top, fixed at the base, with the bracket load applied 0.8 m off the axis.

Approach. Compute the CHS properties and classify it, obtain the compressive resistance $C_r$ with $K = 0.8$ for a fixed–pinned member and the flexural resistance $M_r$, evaluate the moment distribution produced by the bracket eccentricity, then solve the CSA S16 Clause 13.8.3 beam-column interaction equations for the largest $P_F$.

Section properties. With inside diameter $d_i = 406.4-2(9.53) = 387.34$ mm, $$A=\frac{\pi}{4}\left(D^2-d_i^2\right)=\frac{\pi}{4}\left(406.4^2-387.34^2\right)=11\,882\ \text{mm}^2$$ $$I=\frac{\pi}{64}\left(D^4-d_i^4\right)=234\times10^6\ \text{mm}^4,\qquad r=\sqrt{I/A}=140.4\ \text{mm}$$ and the plastic modulus of a circular tube is $Z = (D^3-d_i^3)/6 = 1\,501\times10^3$ mm³.
Classification. For circular hollow sections S16 works with $D/t = 406.4/9.53 = 42.6$. In flexure the Class 1 limit is $13\,000/F_y = 37.1$ and the Class 2 limit is $18\,000/F_y = 51.4$, so the tube is Class 2 and can still develop its plastic moment. In axial compression the limit for a non-slender wall is $23\,000/F_y = 65.7 > 42.6$, so no local-buckling reduction applies.
Compressive resistance. A column fixed at one end and pinned (but held laterally) at the other has a recommended design effective-length factor $K = 0.8$: $$\frac{KL}{r}=\frac{0.8(6\,000)}{140.4}=34.2,\qquad \lambda=\frac{KL}{r}\sqrt{\frac{F_y}{\pi^2E}}=34.2\sqrt{\frac{350}{\pi^2(200\,000)}}=0.455$$ Class H hollow sections use $n = 2.24$ in the S16 column curve: $$C_r=\phi AF_y\left(1+\lambda^{2n}\right)^{-1/n}=0.90(11\,882)(350)(1.0295)^{-0.446}=3\,695\ \text{kN}$$ The cross-sectional (stub-column) value with $\lambda = 0$ is $C_{r0} = \phi AF_y = 3\,743$ kN.
Flexural resistance. A Class 2 closed circular section cannot buckle laterally and reaches its plastic moment: $$M_r=\phi ZF_y=0.90(1\,501\times10^3)(350)=473\ \text{kN}\cdot\text{m}$$
Moment distribution from the bracket. The bracket applies $M = P_Fe$ at the top of a member that is pinned there and fixed at the base — a propped cantilever loaded by an end moment. The internal moment varies linearly, $$M(z)=\frac{M}{2}\left(\frac{3z}{L}-1\right)\ \Rightarrow\ M_{top}=M=0.8P_F,\qquad M_{base}=-\frac{M}{2}$$ so the member bends in double curvature with an end-moment ratio $\kappa = +0.5$, giving $$\omega_1=0.6-0.4\kappa=0.4\ \ (\ge 0.4)$$
Amplification factor. With $K = 1.0$ in the plane of bending, $$C_e=\frac{\pi^2EI}{L^2}=\frac{\pi^2(200\,000)(234\times10^6)}{6\,000^2}=12\,834\ \text{kN}$$ $$U_1=\frac{\omega_1}{1-C_f/C_e}=\frac{0.4}{1-P_F/12\,834}$$ For any $P_F$ near the answer this evaluates to about 0.42, below unity, so the code floor $U_1 = 1.0$ governs — the double-curvature moment diagram is favourable enough that no amplification is required.
Beam-column interaction (S16 Cl. 13.8.3). For a section other than a Class 1 I-shape the interaction has no 0.85 relief factor: $$\frac{C_f}{C_r}+\frac{U_1M_f}{M_r}\le 1.0,\qquad M_f=0.8P_F$$ Overall member strength, using $C_r = 3\,695$ kN: $$P_F\left(\frac{1}{3\,695}+\frac{0.8}{473}\right)=P_F\left(2.71\times10^{-4}+1.69\times10^{-3}\right)\le 1$$ $$P_F\le 510\ \text{kN}$$ Cross-sectional strength, using $C_{r0} = 3\,743$ kN, gives $P_F \le 511$ kN, and because a circular tube has identical properties about every axis the lateral-torsional check reproduces the member check exactly. The smallest value controls: $$\boxed{P_F=509\ \text{kN}}$$
Confirm the bending-only limit. S16 also requires $M_f/M_r \le 1.0$: at $P_F = 509$ kN, $M_f = 0.8(509) = 407$ kN·m and $M_f/M_r = 0.86 < 1.0$, so the answer is not violating the pure-flexure ceiling. Note the split of the interaction: the axial term contributes only 0.14 while flexure contributes 0.86, so this member is effectively a beam with incidental axial load — a longer eccentricity arm would cut the capacity almost in inverse proportion.

Quantity	Value
$A$ / $I$ / $r$ / $Z$	11 882 mm² / 234 × 10⁶ mm⁴ / 140.4 mm / 1 501 × 10³ mm³
$D/t$ → class	42.6 → Class 2 in flexure, non-slender in compression
$KL/r$ ($K$ = 0.8) / $\lambda$	34.2 / 0.455
$C_r$ / $C_{r0}$	3 695 kN / 3 743 kN
$M_r$	473 kN·m
$\omega_1$ / $U_1$ / $C_e$	0.40 / 1.0 (floor) / 12 834 kN
Maximum factored bracket load $P_F$	509 kN

Topic. Beam-column interaction for a hollow structural section.

Key relations. $C_r = \phi AF_y(1+\lambda^{2n})^{-1/n}$ with $n = 2.24$ for Class H HSS; $\lambda = (KL/r)\sqrt{F_y/\pi^2E}$; $M_r = \phi ZF_y$ for Class 1 and 2; $U_1 = \omega_1/(1-C_f/C_e)$ with $\omega_1 = 0.6-0.4\kappa \ge 0.4$; interaction $C_f/C_r + U_1M_f/M_r \le 1$.

Why this works. Axial load and bending interact because the axial force acting through the lateral deflection adds moment — the P-delta effect — and because both actions consume the same yield capacity. S16 handles the first through $U_1$ and the second through the linear interaction sum. Here the bracket moment is applied at a pinned end, so the moment diagram reverses sign inside the member; the resulting double curvature keeps the peak moment and the peak deflection far apart, which is exactly what $\omega_1 = 0.4$ encodes.

Common pitfall. Using $n = 1.34$ instead of 2.24. Class H sections are stress-relieved and carry far smaller residual stresses, so they follow the higher column curve; using 1.34 here would understate $C_r$ by about 5 %. The second frequent error is treating the base moment as $P_Fe$ as well — the propped cantilever redistributes it to $P_Fe/2$ of opposite sign, and the ratio $\kappa$ that follows is what makes $U_1$ fall to its floor.

Source. CSA S16 Clauses 11.2 (Table 2), 13.3.1, 13.5, 13.8.3 and Annex F (effective length); Kulak & Grondin, Limit States Design in Structural Steel, Ch. 8.

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